Is there a better way to find the name of a file than just looping through argv[] looking for an argument that isn't a flag - ?
In this case, the flags can be entered in any order (so optind isn't going to help).
ie:
/program -p file.txt -s
/program -p -s file.txt -b
/program file.txt -p -s -a
int main (int argc, char *argv[]){
char option;
const char *optstring;
optstring = "rs:pih";
while ((option = getopt(argc, argv, optstring)) != EOF) {
switch (option) {
case 'r':
//do something
break;
case 's':
//etc etc
}
}
From the man page of getopt(),
By default, getopt() permutes the contents of argv as it scans, so that eventually all the nonoptions are at the end.
So the order in which the options and non-option arguments are given doesn't matter.
Use getopt() to take care of the options and their arguments (if any). After that, check the value of optind.
As the man page says,
The variable optind is the index of the next element to be processed in argv.
In your command, it seems that there is only one non-option argument. If that's the case after all the options have been correctly processed, optind must be equal to argc-1.
Also, in the optstring you have given, there is colon following s. That means if the option -s is there, it must have an argument.
while ((option = getopt(argc, argv, optstring)) != EOF) {
switch (option) {
case 'r':
//do something
break;
case 's':
//etc etc
}
}
//Now get the non-option argument
if(optind != argc-1)
{
fprintf(stderr, "Invalid number of arguments.");
return 1;
}
fprintf(stdout, "\nNon-option argument: %s", argv[optind]);
Note: The options may be provided in any order but the argument of one option may not be given as the argument of another.
